Let $P$ be an arbitrary point in the interior of triangle $ABC$. Lines $BP$ and $CP$ intersect $AC$ and $AB$ at $E$ and $F$, respectively. Let $K$ and $L$ be the midpoints of the segments $BF$ and $CE$, respectively. Let the lines through $L$ and $K$ parallel to $CF$ and $BE$ intersect $BC$ at $S$ and $T$, respectively; moreover, denote by $M$ and $N$ the reflection of $S$ and $T$ over the points $L$ and $K$, respectively. Prove that as $P$ moves in the interior of triangle $ABC$, line $MN$ passes through a fixed point.
Here's a simple argument using congruence and similarity of triangles and properties of parallel lines:
In $\triangle KBT$ and $\triangle KFN$,
$KB=KF$, $KT=KN$, $\angle TKB=\angle NKF$.
$\therefore \triangle KBT\cong\triangle KFN$.
$\Rightarrow \angle KBT=\angle KFN$
$\Rightarrow BT||FN\Rightarrow BC||FN$.
Let $X$ be the intersection point of lines $BE$ and $NF$.
In $\triangle FBX$, $K$ is the midpoint of $BF$ and $KN||BX$.
$\Rightarrow N$ is the midpoint of segment $FX$.
Now, since $FX||BC$ and $N$ is the midpoint of $FX$, the line $NP$ must meet $BC$ at the midpoint of segment $BC$.
Let's define $Q$ to be the mid point of $BC$. $\therefore N,P,Q$ are collinear.
By similar argument above, we can prove that $M,P,Q$ are collinear.
So, lines $MN$ and $PQ$ are the same line. So, line $MN$ must go through the point $Q$, independent of the position of $P$.
$Q.E.D.$
$M,N,P,Q$ collinear. $Q$ is the midpoint of segment $BC$
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